Find (without using a calculator) the absolute extreme values of the function on the given interval. f(x) = x3 - 9x2 + 15x + 4 on (-1,2] absolute min absolute max Need Help? Read Watch Talk to a Tutor 2. -/10 POINTS BERRAPCALC73.3.011. Find (without using a calculator) the absolute extreme values of the function on the given interval. f(x) = 9x2 - x3 on O, 8] absolute min absolute max Need Help? Read It Watch It Talk to a Tutor

## Answer

At first we find out all possible critical points of the given

function on the given interval. Then we find out absolute maximum

and absolute minimum of the given function on the given interval by

using critical points and end points of the given interval.

so the (un) annat Given function f(n) = 23-94'1154+4 on [1, 2] film) o (m3-91+ 150+4) arom (ny) - pasta () + 15 (Wtest (4) = 3nr-9 (22) + 15 (1) to using = 3nr 18ntis ona ()0 = 3(*v- 61+5) For critical points we get f(n)=0. 7361-6n+5) 20 9 XV-69+5 20 a n-in-uts 20 * n(n-5)-1(4-5) = a R-5) (n-1) zo 1 (n=5) -0 or (x-1) 20 - ax=5 or x=1 Two critical points are x=1 and nas. But x=5 does not lies on [-1,2]. Therefore only critical points of feres on [1, 2] is nal ave

when not (Left end point) . Then ty)=(-1)3-9(-1)+15(-1)+4 2-1-9 -15 +4 3-21 ty) =-2) When k = 1 (Critical point ) Then fci) = 13-91111 +15(1+4 31-9 +15+4 = fru = 11 when se=2 (Right and point) f(3) = 25-9 (27) +15(2) +4 = 8 - 36 + 30 + 4 = 42-36 * 8 a f(2)=6 Therefore, absolute min [21] S absolute man " Giren function f(x) = 96%-n3 on [0,8] f'(X) = ( 911–23) 39.() - 5 (4°) cum) -nunti 39 (2n) - 34 2 34(6-x) using <

[0,8] For critical point we get f'(x)=0 tan (6-4) 20 auzo or 6-12 an26 The critical points are x20 and 126 on when xoo (heft and point also critical point) Then flojz 9(0)7-(0)3 When n=6 (critical point) Then f(6) 2 g (6%-63 2 6²/9-6) = 36 ( 3) 2 108 When nag (Right and point) Then f(8) = g(8)-83 . 764 (1) =64 Therefore, absolute min - absolute mare a [108] man